Inequality and integrals

Question

I want to prove that: $$\int_0^1\frac{2\sin{x}}{x^2+1}\text{d}x\le\ln{2}$$ So I did the followings: $$\int_0^1\frac{2\sin{x}}{x^2+1}\text{ d}x\le\ln{2}\iff\int_0^1\frac{2\sin{x}}{x^2+1}\text{ d}x\le\int_0^1\ln{2}\text{ d}x\iff\int_0^1\frac{2\sin{x}}{x^2+1}-\ln{2}\text{ d}x\le 0$$ Now I want to prove that: $$\frac{2\sin{x}}{x^2+1}-\ln{2}\le 0$$ But I don't know how. Any ideas? (Plus I have prove that $\sin{x}\le x\text{ }\forall\text{ }x\in\text{ }[0,1]$)

Answer 1

As $\sin x\le x$ for $x\in[0,1]$, $$\frac{2\sin x}{x^2+1}\le\frac{2x}{x^2+1}$$ for $x\in[0,1]$.

$$\int_0^1\frac{2\sin x}{x^2+1}dx\le\int_0^1\frac{2x}{x^2+1}dx=\left[\ln(x^2+1)\right]_0^1=\ln 2$$